Are My Free Energy and Equilibrium Constant Calculations Correct?

AI Thread Summary
The calculations for the free energy change associated with the formation of methanol from carbon monoxide and hydrogen gas yield a delta G of -35.92 kJ, indicating a spontaneous reaction under the given conditions. For the formation of compound AB, the equilibrium constant calculated is 0.1018 at 181°C, suggesting a low tendency for the reaction to proceed towards products. Both calculations utilize the standard free energy of formation and the appropriate equations for delta G and equilibrium constant. The values presented appear to be correct based on the provided data and equations. Overall, the calculations for both reactions are valid.
Soaring Crane
Messages
461
Reaction score
0
Are my final values correct?

1. Determine the free energy change associated with the formation of methanol from carbon monoxide and hydrogen gas if the partial pressure of CO is 1.0700 atm, H2 is 2.2800 atm, and CH3OH is 0.0600 atm at 298K. The standard free energy of formation for methanol is - 24.70 kJ. Give your answer in kilojoules.

CO (g) + 2H2 (g) --> CH3OH (g)

deltaG* = -24.70 kJ = -24,700 J

deltaG = G* + RTlnQ

Q = [(CH3O)]/[(CO)*(H2)^2] = [0.06 atm]/[(2.28 atm)^2*(1.07 atm)]
= 0.010786928

delta G = -24,700 J + (298 K)(8.314)ln(0.010786928)
= -35921.965 J
= -35.92 kJ




2. The compound AB can be formed from A and B according to the following reaction:A2 (g) + B2 (g) --> 2 AB (g) If the free energy of formation for AB is 8.66 kJ/mol, what is the equilibrium constant for the reaction at 181C? Give your answer to four decimal places.


deltaG* = 2 mol(8.66 kJ/mol) = 17.32 kJ = 17,320 J

deltaG* = RTlnK

K = e^-(G/RT)
= e^-(17,320 J/(8.314 J *454.15 K))
= e^-(4.5871)
= 0.10182 = 0.1018 ?




Thanks.
 
Physics news on Phys.org
Are my answers valid?

Thanks.
 
Back
Top